The other (http://nathangeffen.webfactional.com/spacetravel/spacetravel.php) says that a 40 kg ship (payload+fuel) making a 1g 4.3 light-year journey with a fuel conversation rate of 1 (which I assume means 100% efficiency) would need 177 kg of fuel on the trip, which is impossible unless it gets fuel during the trip from somewhere.

The first site indicates that with enough fuel, you can have a 1g constant acc/dec trip to any distance, while the second says you cannot even go 1 light-year at 1g acc/dec without your fuel exceeding the mass of your spacecraft. Are these based on different underlying assumptions somewhere? Or am I mistaken in believing that they apply to the same situation? Or is one just wrong?

$\begingroup$Just to clarify: did you calculate the payload as 1kg both times? "A 40kg ship (payload+fuel) [...] would need 177 kg of fuel" is contradictory.$\endgroup$
– AsherMay 29 '16 at 20:24

$\begingroup$The math.ucr.edu site calculations are based off of payload intended to arrive at the destination, while the nathangeffen site calculations are based off of spacecraft mass at launch. So in both situations, there would be a ~40 kg spacecraft at launch. Or if your prefer, the ucr website says that a 38/1 fuel/payload ratio would get you to Alpha Centauri at 1g, which means a 38/39 fuel/launch mass ratio. The second website indicates you would need 177/40 fuel/launch mass ratio, which is impossible unless you start out with ~39 kg fuel and pick up more fuel along the way somehow.$\endgroup$
– user2096078May 29 '16 at 20:35

$\begingroup$There's a "Hollywood" link on the webfactional page: "... Even the most efficient spacecrafts, that convert their fuel mass into energy at the rate of e=mc2 (e.g. by using matter-antimatter annihilation), would always need more than four times as much fuel as the total mass of the spacecraft (including fuel), when they launch, to maintain that acceleration, which is impossible."$\endgroup$
– AsherMay 29 '16 at 20:55

$\begingroup$So similar to the situation I worked out. But if that is true, why is the other website's equation wrong?$\endgroup$
– user2096078May 29 '16 at 21:14

2

$\begingroup$It's worth noting the history of the calculation reproduced at Baez' site. It's been around in various forms since the sci.physics.relativity FAQ on USENET in the pre-web days and has been seen by very many scientists along the way.$\endgroup$
– dmckee♦May 29 '16 at 21:41

2 Answers
2

The maximum velocity $v$, observer time $t$, and traveler proper time $T$ calculated on the "Space travel calculator" site are identical to those on John Baez's relativistic rocket page, but the energy requirement and fuel mass calculations are botched.

If you check the explanation notes provided, the energy requirement is calculated as $e = 2mc^2(\gamma -1)$, where $\gamma = 1/\sqrt{1-v^2/c^2}$ is calculated for the maximum velocity $v$ at the trip midpoint and the mass $m$ is the total initial mass, including both payload and fuel. This is supposed to account for the kinetic energy acquired/dissipated during both acceleration and deceleration, but fails to take into account that at the same time the fuel mass decreases constantly until completely depleted by the end of the trip. With this erroneous assumption in place, the fuel requirement is then calculated as $M = e/rc^2$, where $r$ is the fuel conversion rate or efficiency. So for unit fuel efficiency $r=1$, the fuel mass is incorrectly estimated as
$$
M = 2m(\gamma -1)
$$
where $m$ technically still includes the initial fuel load!

In contrast, the Baez page does account for fuel consumption while applying energy and momentum conservation, and finds that the required fuel (M) to payload (m) mass ratio is
$$
\frac{M}{m} = e^{\frac{aT}{c}} - 1
$$
where $a$ is the constant acceleration in the traveler's frame and $T$ is the total trip proper time. As an exercise, you may want to check that the above expression can be cast in the much simpler form
$$
\frac{M}{m} = \frac{2 \frac{v}{c}}{1 - \frac{v}{c}}
$$
where $v$ is again the maximum velocity at trip midpoint (acceleration and deceleration legs being assumed of practically identical durations).

$\begingroup$I have updated and hopefully corrected the Space Travel Calculator (and the accompanying article on space travel difficulties). Apologies for the incorrect fuel calculation, which sat there for four years! If anyone spots any further bugs, please do let me know.$\endgroup$
– Nathan GeffenJun 5 '16 at 21:13

$\begingroup$Great! Just checked it out, now works as promised :)$\endgroup$
– udrvJun 6 '16 at 2:44

I thought I would indicate how I see this problem. This question involves different ways of defining the payload and total mass of the rocket.

The four momentum of a body in flat spacetime, such as a rocket, is
$$
P~=~(E,~{\bf p}).
$$
The four-momentum has the spatial momentum and the energy. The energy the rocket before it starts to accelerates has the initial energy
$$
E_i~=~(M~+~m)c^2,
$$
where $M$ is the fuel mass and m is the payload mass. Once the fuel, presumably matter plus anti-matter, is used it is converted to photon energy plus the final energy of the system
$$
E_f~ =~\gamma mc^2~+~E_{ph},
$$
where $E_{ph}$ is the energy of the photons generated. Conservation of energy tells us that $E_i~=~E_f$ and so
$$
(M~+~m)c^2~=~\gamma mc^2~+~E_{ph}.
$$
Similarly there is a conservation of momentum. Before accelerating the total spatial momentum is zero in the Earth frame,
$$
P_i~=~0.
$$
After the fuel is converted to energy the final spatial momentum is that of the ship plus that of the photons directed in the opposite direction
$$
P_f~=~\gamma mv~-~E_{ph}/c.
$$
By conservation of momentum $P_f~=~P_i$, and so
$$
\gamma mv~-~E_{ph}/c~=~0.
$$
Eliminating $E_{ph}$ from these two conservation equations gives
$$
(M~+~m)c^2~-~\gamma mc^2~=~\gamma mvc,
$$
and the fuel to payload ratio is then
$$
M/m~=~\gamma(1~+~v/c)~-~1.
$$
The equations for an accelerated reference frame gives
$$
\gamma~=~cosh(gT/c),~v~=~c~tanh(gT/c),
$$
this ratio is then
$$
M/m~=~exp(gT/c)~-~1.
$$
So reach the velocity $v$ under a constant acceleration $g$ this is the required fuel/payload mass ratio. As a practical matter rockets are efficient for this ratio $=~10$ or so. The specific impulse is $s~=~c/g~=~3\times 10^7sec$, and so
$$
s~ln(M/m~+~1)~=~T~=~7.2\times 10^7sec.
$$
So the rocket can accelerate for about two years at a $g~=~10m/sec^2$. Put this into equations for the gamma factor and velocity
$$
\gamma~=~cosh(gT/c)~\simeq~12\\
v~=~tanh(gT/c)~\simeq~.993c,
$$
which is pretty fast. For a low gamma rocket a ratio $M/m~=~2$ will accelerate at $g~=~10m/s^2$ for a proper time
$$
s~ log(M/m~+~1)~=~T~\simeq~3.3\times 10^7sec,
$$
which is approximately $1.05$ year to reach a velocity $v~=~.76c$ at one gee. The Lorentz factor is $\gamma~=~1.5$ that gives the coordinate time on Earth $t~=~(c/g)sinh(gT/c)$ or $t~=~3.88\times 10^{7}sec$ or $1.23 yr$, which is longer than the proper time on the craft. A velocity $v~\simeq~.8c$ is sufficient for sending a probe to a star within a radius of $50ly$.

The equation $M/m~=~exp(gT/c)~-~1$, for the time $T \rightarrow \infty$ just means there is a divergent ratio between the fuel mass and the payloaod and rocket engine mass. Of course this become impractical for large ratios. It is why one would need to increase the acceleration. With more ordinary rockets, it is why we do not build $500$ meter tall gunpowder rocket to launch spacecraft into space, but rather more efficient liquid propelled rockets with higher thrust and specific impulse. For $c/g = 3\times 10^7sec$ this is the maximum specific impulse physically possible. The only thing left is to increase the photon flux out the back to increase thrust.